b. How many milliliters of a 0.8 M solution of citric acid would be needed to react with 15 grams of baking soda?

Question

Chemistry

Shyla Payne

12 February, 19:03

b. How many milliliters of a 0.8 M solution of citric acid would be needed to react with 15 grams of baking soda?

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Answers (1)

Know the Answer?

Douglas Dudley · Answer 1

74.4 ml of millimeters needed to react with 15 grams of baking soda

Explanation:

Given data

n C8 H8 07 = 0.8 M

m NaHCO3=15 g

Solving the problem:

From the equation we get,

3x C6 H8 07 = NaHC03

Finding the molar mass for the baking soda

NaHCO3 = 23+1+12+3 x16 = 84 g / mol

so,

n NaHCO3 = m NaHCO3 / M NaHCO3

And then,

n C6 H8 07 = 1/3 * n NaHCO3

From the definition of the molarity we get

V C6 H8 O 7=n C6 H8 07/C6 H8 07

=n NaHCO3/3 * C6 H8 07 =

m NaHCO3 / M NaHCO3 / 3 * C6 H8 07

Calculating the equation we get,

V C6 H8 O 7 = 15/84/3 * 0.8=5/84 * 0.8 = 0.0744L = 74.4 ml

74.4 ml of millimeters needed to react with 15 grams of baking soda

n C8 H8 07 = 0.8 M

m NaHCO3=15 g

Solving the problem:

From the equation we get,

3x C6 H8 07 = NaHC03

Finding the molar mass for the baking soda

Nahco3 = 23+1+12+3 x16 = 84 g / mol

so,

n NaHCO3 = m NaHCO3 / M NaHCO3

And then,

n C6 H8 07 = 1/3 * n NaHCO3

From the definition of the molarity we get

V C6 H8 O 7=n C6 H8 07/C6 H8 07

=n NaHCO3/3 * C6 H8 07 =

m NaHCO3 / M NaHCO3 / 3 * C6 H8 07

Calculating the equation we get,

V C6 H8 O 7 = 15/84/3 * 0.8=5/84 * 0.8 = 0.0744L = 74.4 ml

74.4 ml of millimeters needed to react with 15 grams of baking soda